# Problem solving an integral

• A
nikos749
TL;DR Summary
i had been trying to solve $$\int_0^{\infty} \frac{e^{iax}-e^{-ibx}}{x^2} dx$$ with mathematica but the result was "Integral does not converge''
I had been trying to aplly the sokhotski–plemelj theorem but with no success.
Moreover i replaced exponential function with taylor expansion but i still can not solve the integral.
thanks

Mentor
2022 Award
I would split it into two integrals, substitute ##x## by ##\dfrac{1}{x}## and look whether I can get it to look like the Gamma function.
$$\int_0^\infty x^{z-1}e^{-\mu x} = \dfrac{\Gamma(z)}{\mu^z} \; , \;Re(z),Re(\mu) > 0$$
$$\int_0^\infty x^{z-1}e^{- i \mu x} = \dfrac{\Gamma(z)}{( i \mu)^z} \; , \;0<Re(z)<1,\mu > 0$$

• dextercioby
nikos749
I would split it into two integrals, substitute ##x## by ##\dfrac{1}{x}## and look whether I can get it to look like the Gamma function.
$$\int_0^\infty x^{z-1}e^{-\mu x} = \dfrac{\Gamma(z)}{\mu^z} \; , \;Re(z),Re(\mu) > 0$$
$$\int_0^\infty x^{z-1}e^{- i \mu x} = \dfrac{\Gamma(z)}{( i \mu)^z} \; , \;0<Re(z)<1,\mu > 0$$
i end up $$\int_0^{\infty} ( e^{ia/x}-e^{-ib/x})dx$$ which is something different from gamma function

Gold Member
I do not understand what is point simplifying it if it is already clear, that the integral does not converge. Even ##\int_0^1(dx*\frac{1}{x^2})## does not converge.

Mentor
2022 Award
I do not understand what is point simplifying it if it is already clear, that the integral does not converge. Even ##\int_0^1(dx*\frac{1}{x^2})## does not converge.
Wolfram Alpha is a good hint, but not a proof. You cannot rule out interpretation errors. E.g. it does converge for ##a=b=0## and Wolfram Alpha didn't consider the possible values of the constants. And what if ##a,b## are multiples of ##2\pi\,?##